FINDING VERTICAL ASYMPTOTES OF TANGENT FUNCTIONS WITH TRANSFORMATIONS

For the function y = tan x, vertical asymptotes are odd multiples of π/2.

We use the characteristics of the tangent curve to graph tangent functions of the form y = A tan (Bx- C), where B > 0

To find the vertical asymptotes of tangent function, we follow the given steps.

Step 1 :

Find two consecutive asymptotes by finding an interval containing one period.

A pair of consecutive asymptotes occurs at 

Vertical asymptote 

for y = tan x is k𝜋 + 𝜋/2 , where k is integer

While equating Bx - C to k𝜋 + 𝜋/2, we will get more asymptotes.

Graph the following tangent function. Find A, the period  and the asymptotes.

Problem 1 :

y = 2 tan (x/2)

Solution :

A = 2

Consecutive asymptotes :

Find two consecutive asymptotes, we do this by finding an interval containing one period.

The period :

An interval containing one period is (-𝜋, 𝜋).

Thus two consecutive asymptotes occur at x = -𝜋 and x = 𝜋.

Finding consecutive asymptotes :

Vertical asymptote of y = tan x is k𝜋 + 𝜋/2, where k is integer. Here

Bx - C = x/2, then

x/2 = k𝜋 + 𝜋/2

x = 2(k𝜋 + 𝜋/2)

x = 2k𝜋 + 𝜋

x = 𝜋(2k + 1)

• When k = -1, x = -𝜋
• When k = 0, x = 𝜋
• When k = 1, x = 3𝜋
• When k = 2, x = 5𝜋

So, the required asymptotes for the given function  are 

-𝜋, 𝜋, 3𝜋, 5𝜋,.............

Problem 2 :

y = tan (x - (𝜋/4))

Solution :

A = 1

Consecutive asymptotes :

Find two consecutive asymptotes, we do this by finding an interval containing one period.

The period :

An interval containing one period is (-𝜋/4, 3𝜋/4).

Thus two consecutive asymptotes occur at x = -𝜋/4 and x = 3𝜋/4.

Finding consecutive asymptotes :

Vertical asymptote of y = tan x is k𝜋 + 𝜋/2, where k is integer. Here

Bx - C = x - (𝜋/4), then

x - (𝜋/4) = k𝜋 + 𝜋/2

x = k𝜋 + 𝜋/2 + (𝜋/4)

x = k𝜋 + 3𝜋/4

x = 𝜋(k + 3/4)

x = (𝜋/4)(4k + 3)

• When k = -1, x = -𝜋/4
• When k = 0, x = 3𝜋/4
• When k = 1, x = 7𝜋/4
• When k = 2, x = 11𝜋/4

So, the required asymptotes for the given function  are 

-𝜋/4,  3𝜋/4, 7𝜋/4, 11𝜋/4, .............

Problem 3 :

y = -tan x - 1

Solution :

y = -(tan x + 1)

A = 1

Since we have negative sign, there should be reflection across y-axis.

Consecutive asymptotes :

Find two consecutive asymptotes, we do this by finding an interval containing one period.

The period :

An interval containing one period is (-𝜋/2, 𝜋/2).

Thus two consecutive asymptotes occur at x = -𝜋/2 and x = 𝜋/2

Finding consecutive asymptotes :

Vertical asymptote of y = tan x is k𝜋 + 𝜋/2, where k is integer. Here

Bx - C = x , then

x = k𝜋 + 𝜋/2

• When k = -1, x = -𝜋/2
• When k = 0, x = 𝜋/2
• When k = 1, x = 3𝜋/2
• When k = -2, x = 5𝜋/2

Asymptotes are all odd multiples 

Problem 4 :

y = tan (x - 𝜋)

Solution :

y = tan (x - 𝜋)

Standard form of tangent function with transformation is 

y = A tan (Bx - C)

A = 1

Consecutive asymptotes :

Find two consecutive asymptotes, we do this by finding an interval containing one period.

The period :

An interval containing one period is (𝜋/2, 3𝜋/2).

Thus two consecutive asymptotes occur at x = 𝜋/2 and x = 3𝜋/2.

Finding consecutive asymptotes :

Vertical asymptote of y = tan x is k𝜋 + 𝜋/2, where k is integer. Here

Bx - C = x , then

x = k𝜋 + 𝜋/2

• When k = -1, x = -𝜋/2
• When k = 0, x = 𝜋/2
• When k = 1, x = 3𝜋/2
• When k = -2, x = 5𝜋/2